a clue to solve $\int ^{1}_{0}(\frac {1-x}{\ln x}\left( x^{2^{0}}+x^{2^{1}}+\ldots +x^{2 ^{n}}\right))dx$

Question

need a clue to solve $$\int ^{1}_{0}(\dfrac {1-x}{\ln x}\left( x^{2^{0}}+x^{2^{1}}+\ldots +x^{2 ^{n}}\right))dx$$ the answer is -ln3.

I have no idea to begin with.

Answer 1

HINT:

$$\int_0^1 x^\text{m}\cdot\frac{1-x}{\ln\left(x\right)}\space\text{d}x=\ln\left(1+\text{m}\right)-\ln\left(2+\text{m}\right)\tag1$$

When $\Re\left(\text{m}\right)>-1$

1. When $\text{m}=2^0$: $$\int_0^1 x^{2^0}\cdot\frac{1-x}{\ln\left(x\right)}\space\text{d}x=\ln\left(1+2^0\right)-\ln\left(2+2^0\right)\tag2$$
2. When $\text{m}=2^1$: $$\int_0^1 x^{2^1}\cdot\frac{1-x}{\ln\left(x\right)}\space\text{d}x=\ln\left(1+2^1\right)-\ln\left(2+2^1\right)\tag2$$

Answer 2

$$\int_{0}^{1}\frac{x^b-x^a}{\ln x}\mathrm dx\\=\int_{0}^{1}\int_{a}^{b}x^u\,\mathrm du\mathrm dx\\ =\int_{a}^{b}\left(\int_{0}^{1}x^u\mathrm dx\right)\mathrm du\\=\int_{a}^{b}\frac{\mathrm du}{u+1}=\ln\frac{b+1}{a+1}$$